Periodic boundary-value problem for third-order linear functional differential equations
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irk-123456789-1644922020-02-10T01:28:43Z Periodic boundary-value problem for third-order linear functional differential equations Hakl, R. Статті 2008 Article Periodic boundary-value problem for third-order linear functional differential equations / R. Hakl // Український математичний журнал. — 2008. — Т. 60, № 3. — С. 413–425. — Бібліогр.: 20 назв. — англ. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/164492 517.9 en Український математичний журнал Інститут математики НАН України |
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Статті Статті Hakl, R. Periodic boundary-value problem for third-order linear functional differential equations Український математичний журнал |
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Periodic boundary-value problem for third-order linear functional differential equations |
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Periodic boundary-value problem for third-order linear functional differential equations |
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Periodic boundary-value problem for third-order linear functional differential equations |
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Periodic boundary-value problem for third-order linear functional differential equations |
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Periodic boundary-value problem for third-order linear functional differential equations |
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periodic boundary-value problem for third-order linear functional differential equations |
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Інститут математики НАН України |
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2008 |
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Periodic boundary-value problem for third-order linear functional differential equations / R. Hakl // Український математичний журнал. — 2008. — Т. 60, № 3. — С. 413–425. — Бібліогр.: 20 назв. — англ. |
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Український математичний журнал |
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AT haklr periodicboundaryvalueproblemforthirdorderlinearfunctionaldifferentialequations |
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2025-07-14T17:02:23Z |
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2025-07-14T17:02:23Z |
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UDС 517.9
R. Hakl (Math. Inst. Acad. Sci. Czech Republic, Brno)
PERIODIC BOUNDARY-VALUE PROBLEM
FOR THIRD ORDER LINEAR
FUNCTIONAL DIFFERENTIAL EQUATIONS*
ПЕРIОДИЧНА ГРАНИЧНА ЗАДАЧА
ДЛЯ ЛIНIЙНИХ ФУНКЦIОНАЛЬНО-ДИФЕРЕНЦIАЛЬНИХ
РIВНЯНЬ ТРЕТЬОГО ПОРЯДКУ
For the linear functional differential equation of the third order
u′′′(t) = `(u)(t) + q(t),
theorems on the existence and uniqueness of a solution satisfying the conditions
u(i)(0) = u(i)(ω), i = 0, 1, 2,
are established. Here, ` is a linear continuous operator transforming the space C
(
[0, ω]; R
)
into the space
L
(
[0, ω]; R
)
, and q ∈ L
(
[0, ω]; R
)
. The question on the nonnegativity of a solution of the considered
boundary-value problem is also studied.
Для лiнiйного функцiонально-диференцiального рiвняння третього порядку
u′′′(t) = `(u)(t) + q(t)
встановлено теореми про iснування та єдинiсть розв’язку, що задовольняє умови
u(i)(0) = u(i)(ω), i = 0, 1, 2.
Тут ` є лiнiйним неперервним оператором, що трансформує простiр C
(
[0, ω]; R
)
у простiр L
(
[0, ω]; R
)
,
а q ∈ L
(
[0, ω]; R
)
. Також розглянуто питання про невiд’ємнiсть розв’язку розглядуваної граничної
задачi.
Introduction. Consider the linear functional differential equation
u′′′(t) = `(u)(t) + q(t), (0.1)
where ` : C
(
[0, ω];R
)
→ L
(
[0, ω];R
)
is a linear continuous operator and q ∈ L
(
[0, ω];R
)
.
By a solution of the equation (0.1) we understand a function u : [0, ω] → R, which
is absolutely continuous together with its first and second derivatives and satisfies the
equation (0.1) almost everywhere in [0, ω].
The following notation will be made use of:
R = ]−∞,+∞[ , R+ = [0, +∞[ ,
[x]+ =
1
2
(|x|+ x), [x]− =
1
2
(|x| − x);
C
(
[0, ω];R
)
is a Banach space of continuous functions u : [0, ω] → R with the norm
‖u‖C = max
{
|u(t)| : t ∈ [0, ω]
}
;
C̃2
(
[0, ω];R
)
is a set of functions u : [0, ω] → R which are absolutely continuous
together with their first and second derivatives;
*The research was supported by the Academy of Sciences of the Czech Republic, Institutional Research
Plan No. AV0Z10190503 and the Grant No. 201/06/0254 of the Grant Agency of the Czech Republic.
c© R. HAKL, 2008
ISSN 1027-3190. Укр. мат. журн., 2008, т. 60, № 3 413
414 R. HAKL
L
(
[0, ω];R
)
is a Banach space of Lebesgue integrable functions p : [0, ω] → R with
the norm
‖p‖L =
ω∫
0
|p(s)|ds;
P is a set of linear nondecreasing operators ` : C
(
[0, ω];R
)
→ L
(
[0, ω];R
)
, i.e.,
such linear operators that `(u)(t) ≥ 0 for t ∈ [0, ω] whenever u(t) ≥ 0 for t ∈ [0, ω].
The equalities and inequalities between functions are understood almost everywhere
in an appropriate interval.
In the present paper, we investigate the question on the existence, uniqueness, and
nonnegativity of a solution of the equation (0.1) satisfying the boundary conditions
u(i)(0) = u(i)(ω), i = 0, 1, 2. (0.2)
The periodic boundary-value problem for higher order ordinary differential equations
has been investigated by many authors (see, e.g., [1 – 11] and references therein). Note
that in [5], unlike the earlier known results, there are investigated, among others, the
existence and uniqueness of an ω-periodic solution of the nonautonomous ordinary
differential equation
u(n) =
n∑
k=1
pk(t)u(k−1) + q(t)
without the requirement on the function p1 to be of constant sign. In this paper we
improve the result of [5] for n = 3 and pk ≡ 0, k = 2, 3, in a certain way (see
Corollary 1.1). For functional differential equations, one can name only a few papers
devoted to the study of the periodic boundary-value problem (see, e.g., [12 – 16]).
The paper is organized as follows. In Section 1, theorems on the existence and
uniqueness, as well as on the nonnegativity, of a solution of (0.1), (0.2) are established.
Sections 2 and 3 are devoted to the proofs of the main results and examples showing
their optimality, respectively.
All the results will be concretized for the differential equation with deviating argument
of the form
u′′′(t) = p(t)u(τ(t)) + q(t), (0.3)
where p, q ∈ L
(
[0, ω];R
)
and τ : [0, ω] → [0, ω] is a measurable function.
Together with the equation (0.1) we will consider the corresponding homogeneous
equation
u′′′(t) = `(u)(t). (0.4)
From the general theory of boundary-value problems for linear functional differential
equations, the following theorem is well-known (see [17], for equations with regular
operators see, e.g., [18, 19]).
Theorem 0.1. The problem (0.1), (0.2) is uniquely solvable if and only if the
corresponding homogeneous problem (0.4), (0.2) has only a trivial solution.
ISSN 1027-3190. Укр. мат. журн., 2008, т. 60, № 3
PERIODIC BOUNDARY-VALUE PROBLEM FOR THIRD ORDER LINEAR FUNCTIONAL ... 415
1. Main results.
Theorem 1.1. Let the operator ` admit the representation ` = `0 − `1, where `0,
`1 ∈ P. Let, moreover, i ∈ {0, 1}, and
ω∫
0
`0(1)(s)ds +
ω∫
0
`1(1)(s)ds 6= 0, (1.1)
ω∫
0
`i(1)(s)ds <
32
ω2
, (1.2)
∫ ω
0
`i(1)(s)ds
1− ω2
32
∫ ω
0
`i(1)(s)ds
≤
ω∫
0
`1−i(1)(s)ds, (1.3)
ω∫
0
`1−i(1)(s)ds ≤ 64
ω2
1 +
√√√√√1− ω2
32
ω∫
0
`i(1)(s)ds
. (1.4)
Then the problem (0.1), (0.2) has a unique solution.
The following two assertions immediately follow from Theorem 1.1.
Corollary 1.1. Let p 6≡ 0, and either
ω∫
0
[p(s)]+ds <
32
ω2
,
∫ ω
0
[p(s)]+ds
1− ω2
32
∫ ω
0
[p(s)]+ds
≤
ω∫
0
[p(s)]−ds ≤ 64
ω2
1 +
√√√√√1− ω2
32
ω∫
0
[p(s)]+ds
,
or
ω∫
0
[p(s)]−ds <
32
ω2
,
∫ ω
0
[p(s)]−ds
1− ω2
32
∫ ω
0
[p(s)]−ds
≤
ω∫
0
[p(s)]+ds ≤ 64
ω2
1 +
√√√√√1− ω2
32
ω∫
0
[p(s)]−ds
.
Then the problem (0.3), (0.2) has a unique solution.
ISSN 1027-3190. Укр. мат. журн., 2008, т. 60, № 3
416 R. HAKL
Corollary 1.2. Let σp(t) ≥ 0 for t ∈ [0, ω], where σ ∈ {−1, 1}, and
0 <
ω∫
0
∣∣p(s)
∣∣ds ≤ 128
ω2
.
Then the problem (0.3), (0.2) has a unique solution.
Remark 1.1. The inequality
ω∫
0
∣∣p(s)
∣∣ds ≤ 128
ω2
in Corollary 1.2 is optimal and it cannot be weakened (see Example 3.1 in Section 3).
Theorem 1.2. Let q(t) ≥ 0 for t ∈ [0, ω] and let the operator ` admit the
representation ` = `0 − `1, where `0, `1 ∈ P. Let, moreover, i ∈ {0, 1}, the conditi-
ons (1.1) – (1.3) be fulfilled, and
ω∫
0
`1−i(1)(s)ds ≤ 32
ω2
. (1.5)
Then the problem (0.1), (0.2) has a unique solution u, and
(−1)iu(t) ≥ 0 for t ∈ [0, ω]. (1.6)
Theorem 1.2 implies the following three assertions.
Corollary 1.3. Let q(t) ≥ 0 for t ∈ [0, ω], p 6≡ 0, and
ω∫
0
[p(s)]+ds <
32
ω2
,
∫ ω
0
[p(s)]+ds
1− ω2
32
∫ ω
0
[p(s)]+ds
≤
ω∫
0
[p(s)]−ds ≤ 32
ω2
.
Then the problem (0.3), (0.2) has a unique solution, and this solution is nonnegative.
Corollary 1.4. Let q(t) ≥ 0 for t ∈ [0, ω], p 6≡ 0, and
ω∫
0
[p(s)]−ds <
32
ω2
,
∫ ω
0
[p(s)]−ds
1− ω2
32
∫ ω
0
[p(s)]−ds
≤
ω∫
0
[p(s)]+ds ≤ 32
ω2
.
Then the problem (0.3), (0.2) has a unique solution, and this solution is nonpositive.
ISSN 1027-3190. Укр. мат. журн., 2008, т. 60, № 3
PERIODIC BOUNDARY-VALUE PROBLEM FOR THIRD ORDER LINEAR FUNCTIONAL ... 417
Corollary 1.5. Let q(t) ≥ 0, σp(t) ≥ 0 for t ∈ [0, ω], where σ ∈ {−1, 1}, and
0 <
ω∫
0
|p(s)|ds ≤ 32
ω2
.
Then the problem (0.3), (0.2) has a unique solution u, and the function σu is nonpositive.
Remark 1.2. The inequality
ω∫
0
∣∣p(s)
∣∣ds ≤ 32
ω2
in Corollary 1.5 is optimal and it cannot be weakened (see Example 3.2 in Section 3).
2. Proofs. To prove Theorems 1.1 and 1.2, we will need the following two lemmas.
Lemma 2.1 can be found in [20] in more general form.
Lemma 2.1. Let u ∈ C̃2
(
[0, ω];R
)
be a nonconstant function satisfying (0.2).
Then
M0 −m0 <
ω2
32
(M2 −m2), (2.1)
where
Mi = max
{
u(i)(t) : t ∈ [0, ω]
}
, mi = min
{
u(i)(t) : t ∈ [0, ω]
}
, i = 0, 2.
Proof. It is obvious that
M2 > 0, m2 < 0, M0 −m0 > 0.
Put
v(t) =
u(t) for t ∈ [0, ω],
u(t− ω) for t ∈ ]ω, 2ω].
Then, obviously, there exist a ∈ [0, ω[ and c ∈ ]a, a + ω[ such that
v(a) = m0, v(c) = M0.
The integration by parts on [a, c] yields
M0 −m0 =
1
2
c∫
a
[
(c− s)(s− a)
]′
v′′(s)ds,
whence, in view of the continuity of v′′, we get
M0 −m0 <
1
2
M2
a+c
2∫
a
[(c− s)(s− a)]′ds + m2
c∫
a+c
2
[(c− s)(s− a)]′ds
=
=
(c− a)2
8
(M2 −m2). (2.2)
ISSN 1027-3190. Укр. мат. журн., 2008, т. 60, № 3
418 R. HAKL
Analogously, on the interval [c, a + ω] we obtain
M0 −m0 <
(a + ω − c)2
8
(M2 −m2). (2.3)
Therefore, multiplying the corresponding sides of the inequalities (2.2) and (2.3), on
account of the fact that 4AB ≤ (A + B)2, we have that (2.1) holds.
The lemma is proved.
Lemma 2.2. Let the operator ` admit the representation ` = `0 − `1, where `0,
`1 ∈ P. Let, moreover, i ∈ {0, 1} and (1.1) – (1.3) be fulfilled. Then every nontrivial
function u ∈ C̃2
(
[0, ω];R
)
satisfying (0.2) and
(−1)iu′′′(t) ≥ (−1)i`(u)(t) for t ∈ [0, ω], (2.4)
assumes positive values.
Proof. Assume on the contrary that there exists a nontrivial nonpositive function
u ∈ C̃2
(
[0, ω];R
)
satisfying (0.2) and (2.4).
First we will show that u is not a constant function. Indeed, supposing u(t) =
= const < 0 for t ∈ [0, ω], the integration of (2.4) from 0 to ω yields
ω∫
0
`1−i(1)(s)ds ≤
ω∫
0
`i(1)(s)ds,
which contradicts (1.1) and (1.3).
Now put
M0 = max
{
|u(t)| : t ∈ [0, ω]
}
, m0 = min
{
|u(t)| : t ∈ [0, ω]
}
, (2.5)
M2 = max
{
u′′(t) : t ∈ [0, ω]
}
, m2 = −min
{
u′′(t) : t ∈ [0, ω]
}
, (2.6)
and choose t1, t2 ∈ [0, ω] such that
u′′(t1) = M2, u′′(t2) = −m2. (2.7)
Obviously,
M0 > 0, m0 ≥ 0, M2 > 0, m2 > 0, (2.8)
and either
t1 < t2 (2.9)
or
t1 > t2. (2.10)
Let (2.9) hold. Then the integration of (2.4) from 0 to t1, from t1 to t2, and from
t2 to ω, respectively, on account of (2.5), (2.7), (2.8), and the assumption `0, `1 ∈ P,
results in
(−1)i(M2 − u′′(0)) ≥ (−1)i
t1∫
0
`(u)(s)ds ≥ −M0
t1∫
0
`i(1)(s)ds, (2.11)
ISSN 1027-3190. Укр. мат. журн., 2008, т. 60, № 3
PERIODIC BOUNDARY-VALUE PROBLEM FOR THIRD ORDER LINEAR FUNCTIONAL ... 419
(−1)i(−m2 −M2) ≥ (−1)i
t2∫
t1
`(u)(s)ds ≥ −M0
t2∫
t1
`i(1)(s)ds, (2.12)
(−1)i(u′′(ω) + m2) ≥ (−1)i
ω∫
t2
`(u)(s)ds ≥ −M0
ω∫
t2
`i(1)(s)ds. (2.13)
If i = 0, then from (2.12) we get
M2 + m2 ≤ M0
ω∫
0
`0(1)(s)ds.
If i = 1, then from (2.11) and (2.13), in view of (0.2), we obtain
M2 + m2 ≤ M0
t1∫
0
`1(1)(s)ds +
ω∫
t2
`1(1)(s)ds
≤ M0
ω∫
0
`1(1)(s)ds.
Consequently, we have
M2 + m2 ≤ M0
ω∫
0
`i(1)(s)ds. (2.14)
Analogously one can show that (2.14) is satisfied also in the case when (2.10) holds.
Thus, in both cases (2.9) and (2.10), the inequality (2.14) holds and, according to
Lemma 2.1, from (2.14) we get
M0 −m0 <
ω2
32
M0
ω∫
0
`i(1)(s)ds. (2.15)
On the other hand, the integration of (2.4) from 0 to ω, by virtue of (0.2), (2.5), and
the assumption `0, `1 ∈ P, yields
m0
ω∫
0
`1−i(1)(s)ds ≤ M0
ω∫
0
`i(1)(s)ds. (2.16)
From (1.1) and (1.3) it follows that
∫ ω
0
`1−i(1)(s)ds 6= 0, and thus (2.15) results in
(M0 −m0)
ω∫
0
`1−i(1)(s)ds <
ω2
32
M0
ω∫
0
`i(1)(s)ds
ω∫
0
`1−i(1)(s)ds. (2.17)
Now, using (2.16) in (2.17), on account of (2.8), we have
ω∫
0
`1−i(1)(s)ds−
ω∫
0
`i(1)(s)ds <
ω2
32
ω∫
0
`i(1)(s)ds
ω∫
0
`1−i(1)(s)ds,
which, in view of (1.2), contradicts (1.3).
The lemma is proved.
ISSN 1027-3190. Укр. мат. журн., 2008, т. 60, № 3
420 R. HAKL
Proof of Theorem 1.1. According to Theorem 0.1 it is sufficient to show that the
problem (0.4), (0.2) has only a trivial solution.
Suppose on the contrary that there exists a nontrivial solution u of (0.4), (0.2). Then,
according to Lemma 2.2, u assumes both positive and negative values. Put
M0 = max
{
u(t) : t ∈ [0, ω]
}
, m0 = −min
{
u(t) : t ∈ [0, ω]
}
, (2.18)
and define numbers M2 and m2 by (2.6). Choose t1, t2 ∈ [0, ω] such that (2.7) holds.
Obviously,
M0 > 0, m0 > 0, M2 > 0, m2 > 0, (2.19)
and without loss of generality we can assume that (2.9) hold.
The integration of (0.4) from 0 to t1, from t1 to t2, and from t2 to ω, respectively,
in view of (2.7), (2.18), (2.19), and the assumption `0, `1 ∈ P, yields
M2 − u′′(0) ≤ M0
t1∫
0
`0(1)(s)ds + m0
t1∫
0
`1(1)(s)ds, (2.20)
M2 + m2 ≤ m0
t2∫
t1
`0(1)(s)ds + M0
t2∫
t1
`1(1)(s)ds, (2.21)
u′′(ω) + m2 ≤ M0
ω∫
t2
`0(1)(s)ds + m0
ω∫
t2
`1(1)(s)ds. (2.22)
If we sum the corresponding sides of (2.20) and (2.22), on account of (0.2), we get
M2 + m2 ≤ M0
∫
I
`0(1)(s)ds + m0
∫
I
`1(1)(s)ds, (2.23)
where I = [0, t1] ∪ [t2, ω]. According to Lemma 2.1, from (2.21) and (2.23) we obtain
M0 + m0 <
ω2
32
(
m0A0 + M0A1
)
, (2.24)
M0 + m0 <
ω2
32
(
m0B1 + M0B0
)
, (2.25)
where
A0 =
t2∫
t1
`0(1)(s)ds, A1 =
t2∫
t1
`1(1)(s)ds, (2.26)
B0 =
∫
I
`0(1)(s)ds, B1 =
∫
I
`1(1)(s)ds. (2.27)
If i = 0, then from (2.24) and (2.25), in view of (1.2) and (2.19), we get
ISSN 1027-3190. Укр. мат. журн., 2008, т. 60, № 3
PERIODIC BOUNDARY-VALUE PROBLEM FOR THIRD ORDER LINEAR FUNCTIONAL ... 421
0 < m0
(
1− ω2
32
A0
)
< M0
(
ω2
32
A1 − 1
)
, (2.28)
0 < M0
(
1− ω2
32
B0
)
< m0
(
ω2
32
B1 − 1
)
, (2.29)
whence we obtain(
1− ω2
32
A0
) (
1− ω2
32
B0
)
<
(
ω2
32
A1 − 1
) (
ω2
32
B1 − 1
)
.
If i = 1, then from (2.24) and (2.25), in view of (1.2) and (2.19), we get
0 < M0
(
1− ω2
32
A1
)
< m0
(
ω2
32
A0 − 1
)
, (2.30)
0 < m0
(
1− ω2
32
B1
)
< M0
(
ω2
32
B0 − 1
)
, (2.31)
whence we obtain(
1− ω2
32
A1
) (
1− ω2
32
B1
)
<
(
ω2
32
A0 − 1
) (
ω2
32
B0 − 1
)
.
Consequently, we have(
1− ω2
32
Ai
) (
1− ω2
32
Bi
)
<
(
ω2
32
A1−i − 1
) (
ω2
32
B1−i − 1
)
. (2.32)
On the other hand, on account of (2.26) and (2.27), using the inequality 4AB ≤
≤ (A + B)2, we find
(
1− ω2
32
Ai
) (
1− ω2
32
Bi
)
≥ 1− ω2
32
ω∫
0
`i(1)(s)ds, (2.33)
(
ω2
32
A1−i − 1
) (
ω2
32
B1−i − 1
)
≤ 1
4
ω2
32
ω∫
0
`1−i(1)(s)ds− 2
2
. (2.34)
Now, using (2.33) and (2.34) in (2.32), we get
1− ω2
32
ω∫
0
`i(1)(s)ds <
1
4
ω2
32
ω∫
0
`1−i(1)(s)ds− 2
2
. (2.35)
Moreover, from (2.28) – (2.31), by virtue of (2.19), (2.26), and (2.27), we have
ω2
32
ω∫
0
`1−i(1)(s)ds ≥ ω2
32
(
A1−i + B1−i
)
> 2. (2.36)
Therefore, the inequality (2.35), on account of (1.2) and (2.36), contradicts (1.4).
The theorem is proved.
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422 R. HAKL
Proof of Theorem 1.2. According to Theorem 1.1, the problem (0.1), (0.2) has a
unique solution u. It is sufficient to show that (1.6) holds.
Suppose the contrary, that there exists t0 ∈ [0, ω] such that (−1)iu(t0) < 0. Then,
according to Lemma 2.2, the function u assumes both positive and negative values.
Define numbers M0, m0, M2, and m2 by (2.18) and (2.6), respectively, and choose
t1, t2 ∈ [0, ω] such that (2.7) is fulfilled. Obviously, (2.19) holds.
Let (2.9) be satisfied. Then the integration of (0.1) from t1 to t2, in view of (2.7),
(2.18), and the assumptions `0, `1 ∈ P and q(t) ≥ 0 for t ∈ [0, ω], results in
m2 + M2 = −
t2∫
t1
(`(u)(s) + q(s))ds ≤
≤ m0
ω∫
0
`0(1)(s)ds + M0
ω∫
0
`1(1)(s)ds.
Let (2.10) be satisfied. Then the integration of (0.1) from 0 to t2 and from t1 to ω,
respectively, in view of (2.7), (2.18), and the assumptions `0, `1 ∈ P and q(t) ≥ 0 for
t ∈ [0, ω], yields
m2 + u′′(0) = −
t2∫
0
(`(u)(s) + q(s))ds ≤
≤ m0
t2∫
0
`0(1)(s)ds + M0
t2∫
0
`1(1)(s)ds,
M2 − u′′(ω) = −
ω∫
t1
(`(u)(s) + q(s))ds ≤
≤ m0
ω∫
t1
`0(1)(s)ds + M0
ω∫
t1
`1(1)(s)ds.
If we sum the corresponding sides of the last two inequalities, on account of (0.2) we
obtain
m2 + M2 ≤ m0
ω∫
0
`0(1)(s)ds + M0
ω∫
0
`1(1)(s)ds. (2.37)
Thus, in both cases (2.9) and (2.10) we have (2.37).
Now, according to Lemma 2.1, the inequality (2.37) results in
M0 + m0 <
ω2
32
m0
ω∫
0
`0(1)(s)ds + M0
ω∫
0
`1(1)(s)ds
,
whence, on account of (1.2) and (1.5), we get a contradiction M0 + m0 < M0 + m0.
The theorem is proved.
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PERIODIC BOUNDARY-VALUE PROBLEM FOR THIRD ORDER LINEAR FUNCTIONAL ... 423
3. Examples.
Example 3.1. Let σ ∈ {−1, 1}, ε > 0 be arbitrary, and choose k ∈ ]0, 1] and
d ∈ ]0, 1/4[ such that 384/((3− 16d2)k) = 128 + ε. Put
x(t) =
96
(3− 16d2)kd
t for t ∈ [0, d[,
96
(3− 16d2)k
for t ∈ [d, 1/2− d[,
96
(3− 16d2)kd
(1/2− t) for t ∈ [1/2− d, 1/2 + d[,
− 96
(3− 16d2)k
for t ∈ [1/2 + d, 1− d[,
96
(3− 16d2)kd
(t− 1) for t ∈ [1− d, 1]
and
u(t) = −(1− t)
t∫
0
sx(s)ds− t
1∫
t
(1− s)x(s)ds.
Then the function u satisfies the conditions (0.2) with ω = 1, and u(3/4) = −u(1/4) =
= 1/k. Therefore we can choose t0, t1 ∈ [0, 1] such that u(t0) = σ, u(t1) = −σ.
Obviously, u is a nontrivial solution of the problem
u′′′(t) = p(t)u(τ(t)), u(i)(0) = u(i)(1), i = 0, 1, 2,
with
p(t) df= σ|x′(t)| for t ∈ [0, ω],
τ(t) df=
t0 for t ∈ [0, 1/4[∪ [3/4, 1],
t1 for t ∈ [1/4, 3/4[.
On the other hand, σp(t) ≥ 0 for t ∈ [0, ω] and
1∫
0
|p(s)|ds = 4 · 96
(3− 16d2)k
= 128 + ε.
Example 3.2. Let σ ∈ {−1, 1}, ε > 0 be arbitrary, and choose k ∈ ]0, 1[ and
d ∈ ]0, 1/4[ such that 96/
(
(3− 16d2)k
)
= 32 + ε. Put
ISSN 1027-3190. Укр. мат. журн., 2008, т. 60, № 3
424 R. HAKL
x(t) =
48
(3− 16d2)kd
t for t ∈ [0, d[,
48
(3− 16d2)k
for t ∈ [d, 1/2− d[,
48
(3− 16d2)kd
(1/2− t) for t ∈ [1/2− d, 1/2 + d[,
− 48
(3− 16d2)k
for t ∈ [1/2 + d, 1− d[,
48
(3− 16d2)kd
(t− 1) for t ∈ [1− d, 1]
and
u(t) = −σ
2
− (1− t)
t∫
0
sx(s)ds− t
1∫
t
(1− s)x(s)ds.
Then the function u satisfies the conditions (0.2) with ω = 1, and u(1/4) = −σ/2 −
− 1/(2k) < −(σ + 1)/2, u(3/4) = −σ/2 + 1/(2k) > (1 − σ)/2. Therefore we can
choose t0 ∈ [0, 1] such that u(t0) = −σ. Obviously, u assumes both positive and
negative values and u is a solution of the problem
u′′′(t) = p(t)u(t0) + q(t), u(i)(0) = u(i)(1), i = 0, 1, 2,
with
p(t) df= σ[x′(t)]−, q(t) df= [x′(t)]+ for t ∈ [0, ω].
On the other hand, q(t) ≥ 0, σp(t) ≥ 0 for t ∈ [0, ω], and
1∫
0
|p(s)|ds = 2 · 48
(3− 16d2)k
= 32 + ε.
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Received 15.10.07
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